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0.25 g of a substance dissolved in 25 g ...

0.25 g of a substance dissolved in 25 g of solvent boiled at a temperature higher by `0.208^@C` than that of the pure solvent. What is the molercular mass of the substance. `[K_(b)" for solvent =2.16 K kg/mol or Km"^(-1)]`

Text Solution

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Given `K_(b)= 2.16K mol^(-1), w= 0.25g, Delta T_(b)= 0.208^(@)C, W=25g`
`Delta T_(b)= (w)/(m xx W)xx 1000 xx K_(b)`
`0.208 =(0.25)/(m xx 25)xx 1000 xx 2.16`
Molar mass of substance, `m= (0.25 xx 1000 xx 2.16)/(0.208 xx 25)= 103.8g mol^(-1)`
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